13.1Twisted conjugacy normal form

Let ℋbe the set of germs along T0 = ?n ×{0} of real analytic functions (Hamiltonians) in ?n × ℝn = {(θ, r)}. The Hamiltonian vector field associated with H ∈ ℋ is

For any given vector α ∈ ℝn, let ?(α) be the affine space of Hamiltonians K ∈ ℋ of the form

for some (nonfixed) c ∈ ℝ; O(r2) stands for the remainder (depending on θ) of the expansion in power of r. The space ?(α) consists exactly of Hamiltonians for which T0 is invariant (r|̇ r=0 = 0) and carries a linear flow with velocity α (θ̇|r=0 = α).24

Let ? be the set of germs along T0 of exact symplectic real analytic isomorphisms of the form25

where φ is an isomorphism of ?n fixing the origin and S is a function of ?n vanishing at the origin. The goal being to find invariant tori close to T0 and carrying a linear flow of frequency α, φ allows us to make changes of coordinates at will on the Lagrangian torus T0, while S allows us to bring back to the zero section any graph over T0, of 0-average and sufficiently close to the zero section.

In the next theorem, we assume that α is Diophantine:

for some fixed γ, τ > 0; we have set |k| = |k1|+⋅ ⋅ ⋅+|kn|. We will call Dγ,τ the set of such vectors. Dγ,τ is nonempty if and only if τ ≥ n − 1 (Dirichlet’s theorem) and, if τ > n − 1 and γ → 0, the complement of Dγ,τ within a ball has measure O(γ); hence, ∪γDγ,τ has full measure [81].

Theorem 1.1 (Herman). If Ko ∈ ?(α) and if H ∈ ℋ is close enough to Ko, there is a unique (K, G, β) ∈ ?(α) × ? ×ℝn such that

We will prove Theorem 1.1 in the next two sections. The statement calls for some remarks.

–The frequency being a conjugacy invariant of quasiperiodic flows, the counterterm β ⋅ r, which allows us to tune the frequency, is necessary. Yet it breaks the dynamical conjugacy between K and H and does not comply H with having an invariant torus, as K does.We call this normal form a twisted conjugacy. The geometrical contents of the theorem are that locally the set of Hamiltonians possessing anα-quasiperiodic torus is a submanifold of finite codimension if α is Diophantine (it has infinite codimension if α is not). The counter-term is the finite-dimensional obstruction to conjugacy to a Hamiltonian of ?(α) and can be imaged as a simple control to preserve a torus of the same frequency and cohomology class as that of Ko.

–In general, one cannot expect H to be of the form

H = (K + β ⋅ r) ∘ G ;

this would show that having a Diophantine invariant torus is an open property,which is wrong, as the following example shows.

Consider the Hamiltonian H = α ⋅ r, α ∈ ℝ2. All the tori r = cst are invariant. Bythe first arbitrarily small perturbation, we may assume that α is resonant: k ⋅α = 0for some k ∈ ℤ2 {0}. Then, add a resonant monomial:

H = α ⋅ r − ϵ sin(2πk ⋅ θ) .

The vector field is

So, the solution through (0, r) at time t = 0 is

t → (tα, r + 2πϵtk) .

So, if ϵ > 0, this solution is unbounded and prevents any invariant torus (among graphs over T0) to exist.

Exercise 1.2. Deduce Arnold’s normal form for vector fields υ on the torus ?n close to a Diophantine rotation [2, 11], from the twisted conjugacy theorem. Hint: Apply the twisted conjugacy theorem to the Hamiltonian H (θ, r) = r ⋅ υ on ?n ×ℝn. Then, check that the zero section is invariant by the corresponding isomorphism G, using an argument of Lagrangian intersection [37].

Bibliographical comments.

13.2One step of the Newton algorithm

Let

We want to solve the following equation between Hamiltonians:

for H close to ϕ(Ko , id, 0) = Ko. The twisted conjugacy theorem, thus, reduces to prove that ϕ is invertible, keeping in mind that

Note that ℋ and ℝn are trivially vector spaces, while

We will invert ϕ using the Newton algorithm, which consists in iterating the operator

Each step of the induction requires to invert the linearized operator ϕ′(x), not only at x0 = (Ko, id, 0), but at some unknown x in the neighborhood of x0, that is, to solve the linearized equation

where δH is the data, (K, G, β) is a parameter, and the unknowns are the “tangent vectors” δK ∈ ℝ⊕ O(r2), δG (geometrically, a vector field along G) and δβ ∈ ℝn. Precomposing with G−1 modifies the equation into an equation between germs along the standard torus T0 (as opposed to the G-dependant torus G−1(T0)):

where we have set G ̇ = δG∘G−1 (geometrically, a germ along T0 of tangent vector field) and H ̇ = δH ∘ G−1. It is a key point in measuring norms that we are interested in the neighborhood of T0 on one side of the conjugacy, and in the neighborhood of G−1(T0) on the other side (Figure 1).

Using the additional notations (in which ∗≥k stands for a function in O(rk ), depending on θ):

and the fact that

and identifying the Taylor coefficients in equation (2.4) yield the following three equations:

The first equation aims at infinitesimally straightening the would-be invariant torusof Ḣ 0, the second equation at straightening its dynamics, and the third at equating higher order terms. Due to the symplectic constraint, the first two equations are coupled (the Hamiltonian lift of the vector field φ̇ having a nontrivial component in the r-direction), whereas in the context of general vector fields, the system is triangular. We will now show the existence of a unique solution to this system of equations, and derive estimates of the solution, within some appropriate functional setting.

Let

be the complex extension of ?n of “width” s, and

for functions f which are real holomorphic on the interior of and continuous on ; such functions form a space which is Banach (there are other possible choices here, e.g., one could consider the space of functions which are real holomorphic on ). We extend this definition to vector-valued functions by taking the maximum of the norms of the components (and, consistently, the ℓ1-norm for “dual” integer vectors, e.g., k ∈ ℤn). Similarly, let be a complex neighborhood of the origin in ℝn of “width” s:

We will call the Banach space of functions which are continuous on and real holomorphic on the interior.

Let

for some fixed CG > 0 and kG > 0 to be determined later.

Lemma 2.1 (Linearized equation). If x is close enough to x0, equation (2.4) possesses a unique solution x ̇ = (δK, Ġ , δβ). Moreover, there exist C′ , τ′ > 0 such that, for all s, σ,

where Cˊ depends only on n, τ, provided K, G−1 and β are bounded on

Proof. First assume that δβ ∈ ℝn is given with |δβ ⋅ r ∘ G−1| ≤ Cst |Ḣ |s+σ, and replace equation (2.6) by

where δβ̂ ∈ ℝn is an additional unknown; as elsewhere in this proof, Cst stands for a constant, to which we do not want to give a consistent name, and which depends only on n, τ, and |(K − α ⋅ r, G−1 − id, β)|s+σ.

–Averagingequation(2.5)yieldsδc = ∫?n (Ḣ 0 + Sˊ ∘ φ−1 ⋅ δβ) dθ; hence,

|δc| ≤ Cst |Ḣ |s+σ.

–According to Lemma B.1, equation (2.5) has a unique solution δS̃ having 0-average, with

Then, the unique solution vanishing at the origin, δS = δS̃ − δS̃(0), satisfies the same estimate (up to an unessential factor 2 which we absorb in the constant).

Note that the estimates hold for all s, σ (at the expense of possibly having an infinite right-hand side). We now proceed similarly with equation (2.9):

–The average yields δβ̂ = ∫?n (Ḣ 1 − 2Ṡˊ ⋅ Q − φˊ ∘ φ−1 ⋅ δβ); hence, using Cauchy’s inequality,26

–The average-free part determines δφ with

Using Cauchy’s inequality and the fact that Qo is given, we see that

where as before the constant depends only on n, τ, and |Qo|s+σ.

Equation (2.7) can then be solved explicitly:

and whence

We have built a map δβ ↦ δβ̂ in the neighborhood of δβ = 0. It is affine and, when φ is close to the identity, invertible. Thus, there exists a unique δβ such that δβ̂ = 0, which satisfies

The claim follows, with τ′ = τc(τc + 1) + 1 and C′ = Cst/γ2 for some constant Cst independent of γ.27

The lemma may be rephrased: the linear operator ϕ′(x) has a unique local inverse ϕ′(x)−1, with the given estimates.

Let

Taylor’s formula says that

where we have set xt = x + t δx (0 ≤ t ≤ 1); hence,

Lemma 2.2 (Remainder). If |Ġ |s+σ ≤ σ/2,

Proof. Let δ2ϕ = ϕ''(K, G, β) ⋅ (δK, δG, δβ)2. We have

hence,

so

note here that δ2ϕ is computed in (K, G, β), and it is then precomposed by G−1.

Now, if xt = (Kt , Gt , βt),

Since

whence the wanted estimate, using (2.10).

It remains to show that the iterated images

of the Newton map (2.2) are defined for n ∈ ℕand converge to some (K, G, β) ∈ ?(α)× ?×ℝn such that H = K ∘ G + β ⋅ r in the neighborhood of G−1(T0), provided H is close enough to Ko. Namely, we will assume that Ko ∈ ?s+σ(α), H ∈ ℋs+σ for some fixed s, σ with 0 < s < s + σ ≤ 1, and

for some ϵ > 0. This is the goal of the next section.

13.3Inverse function theorem

We first give an abstraction of our problem and will afterward show how it allows us to complete the proof of the twisted conjugacy theorem.

Let E = (Es)0<s<1be a decreasing family of Banach spaces with increasing norms |⋅ |s, and = {x ∈ Es , |x|s< ϵ}, ϵ > 0, be its balls centered at 0. Let (Fs) be ananalogous family, and ϕ: → Fs, s < s + σ, ϕ(0) = 0, be maps of Eclass C2, commuting with inclusions.

On account of composition operators, we will assume that there are additional, deformed norms |⋅|x,s, x ∈ Int 0 < s < 1, satisfying

and we will phrase our hypotheses on ϕ in terms of these norms.

Define

Assume that, if x ∈ the derivative Eϕ′(x) : Es+σ → F s has a right inverseϕ′(x)−1 : Fs+σ → Es, and

with C′ , C″, τ′ , τ″ ≥ 1. Let C := C′C″ and τ := τ′ + τ″.

The important fact in the Newton algorithm below is that the index loss σ can be chosen arbitrarily small, without s itself being small, provided the deformed norm substitutes for the initial norm of the spaces F s. The initial norm | ⋅ | s of F s is here only for the practical purpose of having a fixed target space, to which perturbations belong.

Theorem 3.1. ϕ is locally surjective and, more precisely, for any s, η, and σ, with η < s,

In other words, ϕ has a right inverse

Some numbers s, η, and σ, and being given, let

Proof of the theorem. Now, let s, η, and σ be fixed, with η < s and for someϵ.We will see that if ϵ is small enough, the sequence x0 = 0, xn := f n(0) is defined forall n ≥ 0 and converges toward some preimage of y by ϕ.

Let (σn)n≥0 be a sequence of positive real numbers such that 3 Σ σn = σ, and (sn)n≥0 be the sequence decreasing from s0 := s + σ to s defined by induction by the formula sn+1 = sn − 3σn.

Assuming the existence of x0, . . . , xn+1, we see that ϕ(xk) = y + Q(xk−1, xk); hence,

Further assuming that |xk+1 − xk|sk ≤ σk, the estimate of the right inverse and Lemma 10.2 entails that

The estimate

and the fact, to be checked later, that ck ≥ 1 for all k ≥ 0, show

Since Σn≥0 ρ2n ≤ 2ρ if 2ρ ≤ 1, and using the definition of constants ck’s, we get a sufficient condition to have all xn’s defined and to have Σ|xn+1 − xn|s ≤ η:

Maximizing the upper bound of ϵ under the constraint 3 Σn≥0 σn = σ yields σk := A posteriori it is straightforward that |xn+1 − xn|sn ≤ σn (as earlier assumed to apply Lemma 10.2) and cn ≥ 1 for all n ≥ 0. Besides, using that Σ k2−k = Σ2−k = 2, we get

whence the theorem.

Remark. The two competing small parameters η and σ being fixed, our choice of the sequence (σn) maximizes ϵ for the Newton algorithm. It does not modify the sequence (xk) but only the information we retain from (xk).

Exercise 3.2 (End of proof of Theorem 1.1). Complete the proof by checking that

–A similar statement as Theorem 3.1 holds if ϕ is defined only on a ball of polynomial radius with respect to the width of analyticity (see (2.8)).

– and βn are bounded along the induction (in order to justify the repeated use of estimates of Lemmata 2.1 and 2.2,which are not uniform as assumed in (3.1)). Hint: Use the fact that

the estimate of G ̇ in the induction and the estimate of Proposition B.2 in appendix B.

Corollary 3.3. The size of the allowed perturbation is polynomial in the Diophantine constant γ (see (1.1)).

Exercise 3.4. What is the domain of ψ in F S? Hint: Optimize the function ϵ(η, σ) under the constraint s + σ = S.

Bibliographical comments.

13.4Local uniqueness and regularity of the normal form

In the proof of Theorem 3.1, we have built right inverses commuting with inclusions. The proof shows that ψ is continuous at 0; due to the invariance of the hypotheses of the theorem by small translations, ψ is locally continuous.

We further make the following two assumptions:

–Themapsϕ′(x)−1 : Fs+σ → Es are left (as well as right) inverses (in Theorem 1.1 we have restricted to an adequate class of symplectomorphisms).

–The scale(| ⋅ |s) of norms of (Es) satisfies some interpolation inequality:

(according to the sentence after the statement of Corollary C.2 in Appendix C, thisestimate is satisfied in the case it interests to us, since σ + log(1 + σ/s) ≤ σ̃).

Lemma 4.1 (Lipschitz regularity). If σ < s and y, with ϵ = 2−14τC−3σ3τ,

In particular, ψ is the unique local right inverse of ϕ, that is, it is also the local left inverseof ϕ.

Proof. Fix η < ζ < σ < s; the impatient reader can readily look at the end of the proof how to choose the auxiliary parameters η and ζ more precisely.

Let ϵ = 2−8τC−2ζ 2τη, and y, According to Theorem s3.1F, x := ψ(y) and x ̂ := ψ( ŷ ) are in provided the condition, to be checked later, that η < s+ σ−ζ . In particular, we will use a priori that

We have

and, according to the assumed estimate on ϕ′(x)−1 and to Lemma 10.2,

In the norm index of the last term, we will coarsely bound |x̂ − x|s by 2η. Additionally,using the interpolation inequality:

yields

Now, we want to choose η small enough so that

A choice is whence the value of ϵ in the statement.

Proposition 4.2 (Smoothness). For every σ < s, there exists ϵ, C1 such that for every

Moreover, the map defined locally by ψ′(y) = ϕ′(ψ(y))−1 iscontinuous and, if for all σ, so is

Proof. Fix ϵ as in the previous proof and y, Let x = ψ(y), η = y ̂ − y, ξ = ψ(y + η) − ψ(y) (thus, η = ϕ(x + ξ) − ϕ(x)), and Δ := ψ(y + η) − ψ(y) − ϕ′(x)−1η.

Definitions yield

Using the estimates on ϕ′(x)−1 and Q and the latter lemma,

for some σ′ tending to 0 when σ itself tends to 0, and for some C1 > 0 depending on σ. Up to the substitution of σ by σ′, the estimate is proved.

The inversion of linear operators between Banach spaces being analytic, y ↦ ϕ′(ψ(y))−1 has the same degree of smoothness as ϕ′.

Corollary 4.3. If π ∈ L(Es , V) is a family of linear maps, commuting with inclusions,into a fixed Banach space V, then π ∘ ψ is C1 and (π ∘ ψ)′ = π ⋅ (ϕ′ ∘ ψ)−1.

This corollary is used with π : (K, G, β) → β in the proof of Theorem 1.1.

It will later be convenient to extend ϕ−1 to non-Diophantine vectors α. Whitneysmoothness is a criterion for such an extension to exist [94, 98].

Suppose ϕ(x) = ϕα(x) now depends on some parameter α ∈ Bκ (the unit ball ofℝκ),

We will denote ψα the parametrized version of the inverse of ϕα.

Proposition 4.4 (Whitney-smoothness). If s, σ, and ϵ are chosen like in Proposition 4.2,the map is C1-Whitney-smooth and extends to a map of class C1. If ϕ is Ck, 1 ≤ k ≤ ∞, with respect to α, this extension is Ck.

Proof. Let If α, α + β ∈ D, xα = ψα(y) and xα+β = ψα+β(y), we have

Since ŷ ↦ ψα+β(ŷ) is Lipschitz (Lemma 4.1),

and, since α̂ ↦ ϕα̂ (xα) itself is Lipschitz, so is α ↦ xα.

Moreover, the formal derivative of α ↦ xα is

Expanding y = ϕα+β(xα+β) at β = 0 and using the same estimates as above show that

when β → 0, locally uniformly with respect to α. Hence, α ↦ xα is C1-Whitneysmooth, and, similarly, Ck-Whitney-smooth if α ↦ ϕα is.

Thus, by the Whitney extension theorem, the claimed extension exists. Note that Whitney’s original theorem needs two straightforward generalizations to be applied here: ψ takes values in a Banach space, instead of ℝ or a finite-dimensional vector space (see [44]); and ψ is defined on a Banach space, but the extension directions are of finite dimension.

Exercise 4.5 (Quasiperiodic time-dependent perturbations). Let ν ∈ ℝm be fixed. Consider the subspace ℋν of ℋ (in dimension 2(n + m)) consisting of Hamiltonians in

of the form

where H does not depend on Ψ. Since the corresponding Hamiltonian vector field has the component

ℋν may be imaged as the space of Hamiltonians on ?n × ℝn with quasiperiodic time dependence. Show that, if Ĥ ∈ ℋν and Ĥ = ϕ(K, G, (β, β′)) (with β ∈ ℝn and β′ ∈ ℝm), then

Further question: develop the KAM theory below in this particular case.

Exercise 4.6 (Control and persistence of tori). If H is close to an integrable Hamiltonian Ko = Ko(r), show that there is a smooth integrable Hamiltonian β = β(r) for every R such that T0 is a (γ, τ)-Diophantine invariant torus of Ko, H − β(r) ⋅ r has an invariant torus carrying a quasiperiodic dynamics with the same frequency.